--- trunk/doc/cookbook/onedheatdiff001.tex 2009/09/24 00:40:07 2680
+++ trunk/doc/cookbook/onedheatdiff001.tex 2009/09/24 03:04:04 2681
@@ -100,14 +100,14 @@
Neumann boundary conditions describe the radiation or flux normal to the boundary surface. This aptly describes our insulation conditions as we do not want to exert a constant temperature as with the heat source. However, we do want to prevent any loss of energy from the system. These natural boundary conditions can be described by specifying a radiation condition which prescribes the normal component of the flux $\kappa T\hackscore{,i}$ to be proportional
to the difference of the current temperature to the surrounding temperature $T\hackscore{ref}$; in general terms this is;
\begin{equation}
- \kappa T\hackscore{,i} n\hackscore i = \eta (T\hackscore{ref}-T)
+ \kappa T\hackscore{,i} \hat{n}\hackscore i = \eta (T\hackscore{ref}-T)
\label{eqn:hdbc}
\end{equation}
and simplified to our one dimensional model we have;
\begin{equation}
-\kappa \frac{\partial T}{\partial dx} n\hackscore x = \eta (T\hackscore{ref}-T)
+\kappa \frac{\partial T}{\partial dx} \hat{n}\hackscore x = \eta (T\hackscore{ref}-T)
\end{equation}
-where $\eta$ is a given material coefficient depending on the material of the block and the surrounding medium and $n\hackscore i$ is the $i$-th component of the outer normal field \index{outer normal field} at the surface of the domain. These two conditions form a boundary value problem that has to be solved for each time step. Due to the perfect insulation in our model we can set $\eta = 0$ which results in zero flux - no energy in or out - we do not need to worry about the Neumann terms of the general form for this example.
+where $\eta$ is a given material coefficient depending on the material of the block and the surrounding medium and $\hat{n}\hackscore i$ is the $i$-th component of the outer normal field \index{outer normal field} at the surface of the domain. These two conditions form a boundary value problem that has to be solved for each time step. Due to the perfect insulation in our model we can set $\eta = 0$ which results in zero flux - no energy in or out - we do not need to worry about the Neumann terms of the general form for this example.
\subsection{A \textit{1D} Clarification}
It is necessary for clarification that we revisit the general PDE from \refeq{eqn:commonform nabla} under the light of a two dimensional domain. \ESCRIPT is inherently designed to solve problems that are greater than one dimension and so \ref{eqn:commonform nabla} needs to be read as a higher dimensional problem. In the case of two spatial dimensions the \textit{Nabla operator} has in fact two components $\nabla = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y})$. In full, \ref{eqn:commonform nabla} assuming a constant coefficient $A$, takes the form;